Higher Prelim Topic Revision

1 Straight Line

Perpendicular bisectors pass through the midpoint of a line segment, and run perpendicular to them.

\(4y=-3x+29\) or \(y=-\frac{3}{4}x+\frac{29}{2}\)

For more practice, try: Zeta Higher Textbook, Page 15, Exercise 1.8, Questions 1(a), 1(b) and 1(c)

A straight line equation must be rearranged to make \(y\) the subject (\(y=\dots\)) before its gradient can be determined.

\(2y=-x-5\) or \(y=-\frac{1}{2}x-\frac{5}{2}\)

For more practice, try: Zeta Higher Textbook, Page 13, Exercise 1.6, Questions 3, 4 and 5

The formula \(m=\tan{\theta}\) links a line's gradient to the angle it makes with the positive direction of the \(x\)-axis.

\(y=-x+2\)

For more practice, try: Zeta Higher Textbook, Page 9, Exercise 1.4A, Questions 1, 2 and 3

Calculate the gradients of PQ and QR first. The conclusion to a collinearity problem needs to be learned carefully.

\(m_{PQ}=m_{QR}=-3\), valid statement

For more practice, try: Zeta Higher Textbook, Page 12, Exercise 1.5, Questions 1(a), 1(b) and 1(c)

Medians meet opposite sides at the midpoint.
Altitudes meet opposite sides at an angle of 90 degrees.
You can use \(y=y\) as a first step to finding points of intersection.

\(y=-3x+5\)
\(2y=-x-5\) or \(y=-\frac{1}{2}x-\frac{5}{2}\)
\((3,-4)\)

For more practice, try: Zeta Higher Textbook, Page 21, Exercise 1.11B, Questions 1, 2 and 3

2 Recurrence Relations

Calculate \(u_1\) first using the value of \(u_0\) as a starting point.
What is the value of \(a\)?

What is the formula for the limit of a recurrence relation?

\(15\)
\(-1<a<1\), valid statement

\(9\)

For more practice, try: Zeta Higher Textbook, Page 90, Exercise 6.1, Questions 1(a), 1(b) and 1(c)

Can you put both \(u_2\) and \(u_3\) into the recurrence relation? Where does each one go?
What is the value of \(a\)?

\(-4\)
\(a>1\), valid statement

For more practice, try: Zeta Higher Textbook, Page 93, Exercise 6.3, Questions 1(f), 1(i) and 1(l)

Substitute the value of \(u_4\) into the recurrence relation.
Use your expression from part (a), and the new information that \(u_5=-1\).

\(6k-4\)
\(k=\frac{1}{2}\)

For more practice, try: Zeta Higher Textbook, Page 15, Exercise 1.8, Questions 1(a), 1(b) and 1(c)

What "12% is subtracted (from the original 100%) as a decimal?
Since dividing by \(0.12\) is tricky, multiply both parts of the fraction by \(100\) and work from there.

\(a=0.88,b-30\)
250 squirrels

For more practice, try: Zeta Higher Textbook, Page 96, Exercise 6.5B, Questions 1, 2 and 3

Create two equations: one linking the first two terms, and a second linking the second term and the third term. Solve simultaneously.

\(p=-\frac{1}{2},q=2\)

For more practice, try: Zeta Higher Textbook, Page 91, Exercise 6.2, Questions 1(a), 1(b) and 1(c)

3 Differentiation I

To differentiate, multiply by the power and then reduce the power by 1...

\(-19\)

Start by splitting the fraction into two fractions, each with a denominator of \(3x^2\), and prepare each part for differentiation.

\(2x^2+2x^{-3}\)

The gradient of a tangent to a curve is described by its derivative, \(\dfrac{dy}{dx}\).

\(9\)

The gradient of a tangent was found in the last question. To find its equation, we also need to know a coordinate the tangent passes through.

\(y=6x-14\)

The rate of change of a function is described by its derivative, \(h'(t)\). A square root \(\sqrt{x}\) can also be written in index as \(x^{\frac{1}{2}}\).

\(2\)

Find the derivative, \(\dfrac{dy}{dx}\), first.

If a value of \(x\) is substituted into the derivative, the result will be the gradient. Here, when \(x=2\) is substituted in, what value would the output be equal to?

\(k=-\frac{1}{2}\)

4 Quadratic Theory

A few methods may be used.

One begins by fully expanding \(p(x+q)^2+r\).

\(3(x-4)^2+9\)

A few methods may be used.

One begins by fully expanding \(a(x+b)^2+c\).

\(-(x-2)^2+7\)

Start by considering the roots of the equation \(2x^2+8x-10\). Notice the common factor.

You need to draw a sketch.

\(x<-5,x>1\)

Start by considering the roots of the equation \(m^2-m-20\).

You need to draw a sketch.

Since \(m\) is the variable, your answer must be in terms of \(m\).

Coming soon.

An equation has (real) equal roots when \(b^2-4ac=0\).

\(p=\pm6\)

An equation has no real roots when \(b^2-4ac<0\).

\(q<-\frac{6}{5}\)

An equation has real, distinct roots when \(b^2-4ac>0\).

\(k<-2,k>6\)

Points of intersection can be found by solving \(y=y\)...

\((-2,1)\) and \((4,5)\)

Tangency can be explored by solving \(y=y\) are for generally finding points of intersection, then considering how the resulting equation can be factorised or \(b^2-4ac\).

\((2,5)\)